題目

Problem

4. Let $a > 0$ be a constant. Evaluate $\int_0^a \frac{x^2}{(x^2 + a^2)^{3/2}}\,\mathrm{d}x. \quad (10\%)$

解答

展開

根式項含有 $x^2 + a^2$ ，採用正切代換：令 $x = a\tan\theta \implies \mathrm{d}x = a\sec^2\theta\,\mathrm{d}\theta$ 。
轉換積分範圍：
- 當 $x = 0 \implies \theta = 0$
- 當 $x = a \implies \theta = \frac{\pi}{4}$
將代換式代入，進行三角函數化簡後求解定積分。

展開

令 $x = a\tan\theta \implies \mathrm{d}x = a\sec^2\theta\,\mathrm{d}\theta$ 。分母項為：

(x^2 + a^2)^{3/2} = (a^2\tan^2\theta + a^2)^{3/2} = (a^2\sec^2\theta)^{3/2} = a^3\sec^3\theta

積分限轉換：

代入積分式：

I = \int_0^{\frac{\pi}{4}} \frac{a^2\tan^2\theta}{a^3\sec^3\theta} \cdot a\sec^2\theta\,\mathrm{d}\theta

= \int_0^{\frac{\pi}{4}} \frac{\tan^2\theta}{\sec\theta}\,\mathrm{d}\theta = \int_0^{\frac{\pi}{4}} \frac{\sin^2\theta}{\cos\theta}\,\mathrm{d}\theta

= \int_0^{\frac{\pi}{4}} \frac{1 - \cos^2\theta}{\cos\theta}\,\mathrm{d}\theta = \int_0^{\frac{\pi}{4}} (\sec\theta - \cos\theta)\,\mathrm{d}\theta

= \Big[ \ln|\sec\theta + \tan\theta| - \sin\theta \Big]_0^{\frac{\pi}{4}}

= \left( \ln|\sqrt{2} + 1| - \frac{\sqrt{2}}{2} \right) - (\ln|1 + 0| - 0)

= \ln(\sqrt{2} + 1) - \frac{\sqrt{2}}{2}

結論： 定積分值為 $\displaystyle \ln(\sqrt{2} + 1) - \frac{\sqrt{2}}{2}$ 。